In the first answer to this question the users states that if we define a norm $$\|f\|_{k,p} = \left( \sum_{|\alpha| \leq k} \|D^{\alpha} f\|_p^p \right)^{1/p},$$ and write $$\widetilde{C}^k(\Omega) = \{f \in C^k(\Omega) : \|f\|_{k,p} < \infty\},$$ then $W^{k,p}(\Omega)$ is the completion of $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$ as long as $p \in [1, \infty)$.
Why is $W^{k,p}(\Omega)$ the completion of $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$? I don't see the difference between them. What functions are missing from $(\widetilde{C}^k(\Omega), \|\cdot\|_{k,p})$ that are in $W^{k,p}(\Omega)$?
The space $\tilde{C}^k$ is not complete with respect to the norm $||.||_{k,p}$ -- assuming the norm $||.||_p$ is the $L^p$ norm $$||f||_p^p := \int_\Omega |f|^p dx$$ The reason is that convergence in this kind of norm does not preserve differentiability.
Edit: provide simple example to reply to a comment: look, for example, at the functions $$f_n(x) = |x|^{1+\frac{1}{n}}$$ defined on the interval $(-1,1)$. These funktions are (once) continously differentible, and it is not difficult to see that they all belong to what you called $\tilde{C}^1((-1,1))$ It's also not difficult to see that they converge (pointwise and uniformly, first of all) to the function $f(x) = |x|$ They do form a Cauchy Sequence in that space with respect to the $||.||_{1,p}$-norm, too (for each $p$, which I leave as an exercise to you to show), but the limit (which can only be $f$) does not belong to that space. So convergence in that norm does not preserve differentiability.
It is in fact known that $W^{1,1} $ is the space of absolutely continuous functions, which is much larger thant the space of continuous differentiable funcions.